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Hazards of Radiation Radioactivity.

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1 Hazards of Radiation Radioactivity

2 Learning Objectives Recap properties of α, β, γ radiation.
Describe and explain the relative hazards to humans when exposed to α, β, γ radiation. Recap the inverse square law of radiation. Describe how this applies to the safe handling of radioactive sources.

3 Quick Recap Complete the table (in pencil) that describes the properties of the three common radiations:- Radiation Particle Range in air Stopped By

4 Recap - Answers Radiation Particle Range in air Stopped By Alpha
Helium nucleus (P) Few mm (P) Paper (P) Beta High speed electron (P) Few cm (P) Aluminium sheet (P) Gamma Energetic photon (P) Infinite (P) Several cm lead (P)

5 Ionisation What is ionisation?
Which do you think is the most ionising?

6 Ionisation α - Intense, ionises about 10,000 atoms per alpha particle. (because strongly +ve and large) β - Less intense than a, about 100 ionisations per particle. (because lower charge, but higher speed) γ - Weak interaction about 1 ionisation per mm.

7 Hazards α – Once it gets in it is highly damaging to body tissue.
Luckily can’t pass through skin but could be inhaled or ingested. Considered less damaging than gamma rays or alpha particles. β – Lower interaction rate means it is much less damaging to body tissue than alpha. Used as medical tracers. γ - can be a dangerous form of radiation, as they are very penetrating.  Needs intense or prolonged radiation to cause damage to cells.

8 Annoying Question (10 mins)
Explain the dangers associated with radioactive sources (8 marks). Write a mark scheme for this question i.e. assign 1 mark each to 8 short but relevant statements.

9 Possible Mark Scheme Alpha is intensely ionising.
Although short range and kept out by skin Ingestion of alpha emitters can do immense damage to the cells. Beta is less ionising But can penetrate the body Gamma is highly penetrating But causes little ionisation. Long term exposure leads to damage to DNA

10 Spare Question Alpha and beta particles lose about 5 × J of kinetic energy in each collision they make with an air molecule.  An alpha particle makes about 105 collisions per cm with air molecules, while a beta particle makes about 103 collisions.  What is the range of an alpha particle and a beta particle if both start off with an energy of 4.8 × J?

11 Spare Question Answer Both particles lose all their energy in 4.8 × J ÷ 5 × J (P)   = collisions (P)   The alpha particle has a range of ÷ 105 cm-1 = 0.96 cm (P)   The beta particle has a range of ÷ 1000 cm-1 = 96 cm. (P)

12 Inverse Square Law for γ
When taking a reading of intensity, I, at a distance x from the source, I0 is the intensity at the source and k is a constant. Intensity (number of photons per unit area) decreases by the square of the distance. So doubling distance from source means only a quarter of radiation reaches you – so keep them at a distance! -tongs Can be written as:- or:-

13 Learning Objectives Recap properties of α, β, γ radiation.
Describe and explain the relative hazards to humans when exposed to α, β, γ radiation. Recap the inverse square law of radiation. Describe how this applies to the safe handling of radioactive sources.

14 Nuclear Radius Nuclear Physics

15 Homework Research and explain how electron diffraction can be used to determine the radius of the nucleus (6 Marks) Past Paper Question on today’s material. Complete both by Next Lesson, next Monday, Period 5.

16 Learning Objectives State and use the equation for dependence of radius on nucleon number. Calculate nuclear density. Recall the implications of the high nuclear density compared to atomic density.

17 Equation Dependence of radius on nucleon number:-
[The term A1/3 means the cube root of A, the nucleon number.  The term r0 is a constant with the value 1.4 × m.  R is the nuclear radius.] What physical quantity is r0? Rearrange in the form of A=. Try working out R for Gold (A =197 ) and Carbon (A=12)

18 Nuclear Density Radius of a carbon nucleus ~ 3.2 x 10-15m.
Radius of a gold nucleus ~ 8.1 x 10-15m. Mass of a carbon nucleus ~ 2.00 x 10-26kg. Mass of a gold nucleus ~ 3.27 x 10-25kg. What are the densities of the nuclei?

19 Nuclear Density Density of carbon nucleus ~ 1.46 x 1017 kg m-3.
Density of gold nucleus ~ 1.47 x 1017 kg m-3. Very high! One teaspoon = 500 million tonnes. So pretty much the same, regardless of element. Ext: Work out mass of neutron star based on this density. How does it compare to solar mass?

20 Nuclear Density Nuclear density >> Atomic Density This implies:-
Most of an atom’s mass is in its nucleus. The nucleus is small compared to the atom. An atom must contain a lot of empty space.

21 Example Exam Questions
(a)If a carbon nucleus containing 12 nucleons has a radius of 3.2 x 10-15m, what is r0? (b) Calculate the radius of a radium nucleus containing 226 nucleons. (c) Calculate the density of a radium nucleus if its mass is 3.75 x kg. Q2: A sample of pure gold has a density of kg m-3. If the density of the gold nucleus is 1.47 x 1017kg m-3 discuss what this implies about the structure of a gold atom.

22 Learning Objectives State and use the equation for dependence of radius on nucleon number. Calculate nuclear density. Recall the implications of the high nuclear density compared to atomic density.

23 Nuclear Radius Probing Matter

24 Homework Why are spent fuel rods more radioactive after removal from the reactor? What are the different types of radioactive waste? How is each type stored/disposed?

25 Filling the Gaps γ emitters can indeed be used as medical tracers as we thought. Cell membranes can also be destroyed as well as affecting DNA.

26 Learning Objectives Explain how an estimate for the nuclear radius can be obtained from Rutherford’s Experiment.

27 Rutherford’s Alpha Scattering
At P, the point of closest approach, all of the initial kinetic energy of the alpha is converted to electrostatic potential energy.

28 Electrostatic Potential Energy
The equation for Electrostatic Potential Energy is given by this equation:- Where:- EP is the Electrostatic Potential Energy rc is the distance of closest approach ε0 is the permittivity of free space (constant - see data booklet) Q1Q2 are the charges of the two particles involved.

29 Little bit of Maths… Solving to find rC, what do we get? Rearranging…
Remember: Q1=2e = 2 × 1.60 × C (α is He nucleus) Q2=79e = 79 × 1.60 × C (79 protons in Gold nucleus) EP=7.68 MeV = 768 × 106 × 1.60 × J (K.E. of α particles fired at the foil.) ε0 = 8.85 × Fm-1 (from the data booklet) rc= 2.96 × m (a bit large)

30 A couple of points… The nucleus is treated as a point charge.  At this level it is not. The alpha particles are stopped some distance away from the nucleus. It takes higher energy alpha particles to penetrate the nucleus. The values for the nuclear radius given by other particles such as protons, neutrons and electrons are slightly different. …so really it is only an upper limit on the nuclear radius.

31 Method 2 About 1 in 10,000 particles are deflected by more than 90º.
For a thin foil (so that only one scattering) with n layers of atoms, the probability of being deflected is about 1 in 10,000n. This probability depends on effective cross section of nucleus to the atom:- Typically n=10,000 so d = D/10,000

32 Questions A) For a metal foil which has layers of atoms, explain why the probability of an α particle being deflected by a given atom is therefore about 1 in 10,000n. (assume 1 in deflected by more than 90º) B) Assuming this probability is equal to the ratio of cross sectional area of the nucleus to that of the atom, estimate the diameter of a nucleus for atoms of diameter 0.5 nm in a metal foil of thickness 10 μm.

33 Safety Aspects Nuclear Energy

34 Physics Workshop Every Wednesday 3.40pm-4.40pm in O8
Is this time good for most people?

35 Learning Objectives Describe the range of safety features associated with a nuclear reactor. Derive the equation relating half life and the decay constant. Practice calculations on radioactive decay.

36 High Energy Electron Diffraction
Any 6 from:- A beam of high energy electrons is directed at a thin sheet of an element and accelerated through a potential difference of about 108 volts (MeV/high energy) A detector measures the number of electrons diffracted at a number of different angles. Scattering effects occur due the charge of the nuclei and electron and this causes the count rate of the beam of electrons to decrease as angle increases. The electrons are also diffracted by the nuclei in the sheet which causes minima and maxima to observed in the final pattern as long the de Broglie wavelength of the electrons is of the same order as the size of the nucleus, which is about m. The diameter of the nucleus can be calculated using the angle to the first minimum, θmin and the wavelength of the incoming beam λ using the diffraction equation R sin θmin = 0.61λ.

37 Safety Features The reactor is a thick steel vessel designed to withstand the high pressure and temperature in the core. The core is in a building with very thick concrete walls which absorb the neutrons and gamma radiation Emergency shut down system – designed to insert the control rods fully into the core. Sealed fuel rods are inserted and removed using remote handling devices. Spent rods are more radioactive than before use.

38 How to Remember? C - Concrete building. R - Remote Handling.
E - Emergency shut down system. S - Steel vessel for reactor core. S - Spent fuel rods much more radioactive. Spells CRESS, helps us remember?

39 Dangers of Nuclear Power
Chernobyl (1986) Wanted to see if coolant pumps would keep operating if there was a loss of power. When they pushed control rods into reactor, caused loss of power and reversed direction of the rods!

40 Aftermath Overheating caused decomposition of water into hydrogen and oxygen which gases collected at the top of the vessel. Ignited and blew the lid off the reactor and turned the vessel on its side. Nine tonnes of caesium-137 floated across Europe along with many other tons of radioactive material. Caesium-137 is water-soluble and extremely toxic in minute amounts (half life 30 years).

41 Okay let’s try it Describe the range of safety features associated with a nuclear reactor (5 marks).

42 Specific Heat Capacity & Latent Heat - Theory
Thermal Physics Lesson 1

43 Learning Objectives Define specific heat capacity
Perform calculations using ∆Q=mc ∆θ Describe how specific heat capacity can be measured in the lab

44 Homework Complete worksheet to practice calculations we will look at today.

45 Why do we care The anomalously large SHC of water is particularly important for the development and maintenance of life on Earth.

46 Why are we learning this?
Historical Context Images of Joule, Kelvin Steam engines Thermal physics in stars

47 Which is hotter. Bath water or a lit match
Which is hotter? Bath water or a lit match? Which needed more energy to heat it?

48 Heat vs. Temperature Heat as water and temperature as wetness analogy
‘Heat’ is not an entity but a short hand name for a process (heating as oppose to working).

49 State vs. Phase Solids, liquids and gases are three of the different phases of matter (superfluids and plasmas are two others. Thus melting, boiling etc are changes of phase. Each phase can exist in a variety of states depending upon e.g. the temperature and pressure.

50 Specific Heat Capacity
Definition:- The specific heat capacity (c) of a substance is the amount of energy needed to raise the temperature of 1kg of the substance by 1K (or 1 ºC)

51 Specific Heat Capacity
Equation: where:- ∆Q is the energy change in J m is the mass in kg c is the specific heat capacity in J K-1 kg-1 ∆θ is the temperature change in K

52 Worked Example 1 A bucket containing 11.5 litres of cold water at 10°C is taken into a house at a warmer temperature and left inside until it has reached thermal equilibrium with it new surroundings. If 504 kJ of energy is absorbed from the surroundings to heat the water, what is the temperature of the room? ∆Q =504 kJ m=11.5 kg c = 4200 J kg-1 °C-1 ∆θ = ? Using ∆ θ = 504,000 J / (11.5kg × 4200 J K-1 °C-1 ) = 10.4 °C (to 3 s.f.) So temperature of the water = 10°C °C = 20.4 °C Temperature of the room = T of water in thermal eqm. = 20.4 °C

53 Worked Example 2 A 60 W immersion heater takes 2.5 minutes to heat 0.5 kg of water from 21 to 25, what is the specific heat capacity of the water. P = 60 W ∆ t = 2.5 minutes = (2.5 × 60)s = 150 s ∆θ = θ2 - θ1 = 4 oC = 4 K m=0.5 kg c =? ∆Q = P ∆ t = 60 W × 150 s = 9,000 J c = 9,000 J / (0.5 kg × 4 K) = 4,500 J kg-1 K-1 Why is it higher than the accepted value of c = 4200 J kg-1 K-1?

54 Note If a question includes heating and then melting a substance then use:-

55 Learning Objectives Define specific latent heat
Perform calculations using ∆Q=ml Describe how specific latent heat of fusion/vaporisation can be measured in the lab Explain why energy is needed to evaporate a liquid/melt a solid

56 Safety Students must not sit down to watch this experiment - serious scalding has occurred when the beaker breaks or falls and the pupil has been unable to move away instantly.

57 Demonstration What tells you the water is boiling?
So energy is being supplied but the temperature is not rising? What is going on? Work is being done to separate the particles against intermolecular attractive forces.

58 Heating/Cooling Curves
Remember from GCSE?

59 s.h.c vs. latent heat (simple terms)
Specific Heat Capacity Energy needed to heat something Latent Heat Energy needed to change phase

60 Definition The specific latent heat (l) of fusion or vaporisation is the quantity of thermal energy required to change the state of 1kg of a substance. Fusion (liquid  solid) Vaporisation (liquid  gas) …or the other way (Melting/Condensing)

61 Equation where:- ∆Q is the energy change in J
m is the mass of substance changing phase in kg lv is the latent heat of vaporisation in J kg-1 lf is the latent heat of fusion in J kg-1

62 Worked Example 1 The specific latent heat of fusion (melting) of ice is 330,000 J kg-1. What is the energy needed to melt 0.65 kg of ice? ∆Q = ml = 0.65 kg × 330,000 J kg-1 = 210,000 J (2 s.f.)

63 (
Worked Example 2 The power of the immersion heater in the diagram is 60 W. In 5 minutes, the top pan balance reading falls from 282g to 274g. What is the specific latent heat of vaporisation of water? P = 60 W ∆ t = 5 minutes = (5 × 60)s = 300 s m = m2 - m1 = 282g – 274g = 8g = kg lv = ? ∆Q = P∆ t = 60 W × 300s = 18,000 J lv = ∆Q/m = 18,000 J / kg = 2.3 × 106 J kg-1 (2 s.f.) (

64 Gas Laws & Absolute Zero
Thermal Physics Lesson 3

65 Learning Objectives State the three gas laws, describing the relationships between p,V,T and mass Describe how the thermodynamic scale is defined. Define absolute zero. Convert temperatures between Celsius and Kelvin.

66 Thermodynamic Temperature Scale
A change of 1 K equals a change of 1 oC. To convert from degrees Celcius into kelvin add :-

67 Definition Absolute Zero is the lowest possible temperature, and something at this temperature has the lowest possible internal energy. This is zero kelvin, written 0 K, on the thermodynamic temperature scale.

68 Boyle’s Law Temperature and moles of gas are constant
Graph is hyperbolic (see below) and asymptotic to both axes Pressure and volume are inversely proportional to each other

69 Ideal Gas vs. Perfect Gas
Strictly speaking an ideal gas is one that obeys Boyle’s law with complete precision. A perfect gas is a real gas under conditions that Boyle’s law is a valid enough description of its behaviour.

70 Charles’ Law Pressure and moles of gas are constant
Graph is linear (see right) Volume and temperature are directly proportional to each other

71 Pressure Law (Gay-Lussac's Law )
Volume and moles of gas are constant Graph is linear (see right) Pressure and temperature are directly proportional to each other

72 Avogadro’s Law The most significant consequence of Avogadro's law is that the ideal gas constant has the same value for all gases. This means that the constant is given by:-

73 Avogadro’s Constant One mole of any gas contains the same number of particles. This number is called Avogadro’s constant and has the symbol NA. The value of is 6.02 × 1023 particles per mole.

74 Ideal Gas Equation/Molar & Molecular Mass
Thermal Physics Lesson 5

75 Learning Objectives Perform calculations using the ideal gas equation pV=nRT Describe the conditions for which a real gas behaves like an ideal gas Define a mole Calculate the number of moles in a gas using N=nNA

76 The Ideal Gas Equation Combining the three gas laws gives the equation:- The constant is equal to nR. Works well for gases at low pressure and fairly high temperatures

77 Equation of State Recall that each phase can exist in a variety of states e.g. the temperature and pressure Thus the Ideal Gas Equation of State pV = nRT summarises the physically possible combinations of p, V and T for n moles of the ideal gas.

78 Kinetic Theory of Gases
Thermal Physics

79 Learning Objectives Explain the increase of pressure of a gas when it is compressed or heated. Describe the distribution of molecular speeds. Derive the kinetic theory equation.

80 Explaining the gas laws
Assume a gas consists of point molecules moving about at random, continually colliding with the container walls. Each collision causes a force on the container and it is the force of these many impacts that causes the pressure of the gas on the walls.

81 Explaining the gas laws
Boyle’s Law – reducing volume means less distance between collisions with the walls so more collisions per second. Pressure Law- raising the temperature increases the average speed of the molecules so more frequent and harder impacts with the walls.

82 Molecular Speeds The molecules in a gas have a range of speeds.
The root mean square speed, crms, of the molecules is given by:- Where N is the number of molecules in the gas and c1,c2 are the speeds of molecules 1,2, and so on. Note that this is not the same as mean speed.

83 Molecular Speeds

84 The Kinetic Theory Equation
For an ideal gas consisting of N identical molecules, each of mass m, in a container of volume V, the pressure p of the gas is given by:- Where crms is the root mean square speed of the gas molecules. This is the equation we are going to derive.

85 First, some assumptions
Volume of the molecules negligible compared with volume of the gas. They do not attract each other. They move in continual random motion. The collisions are elastic collisions (no loss of k.e.) Duration of collisions much shorter than time between collisions.

86 Remembering Assumptions
R - Random motion. A – Do not Attract each other. V - Volume of the molecules is negligible E - Elastic collisions D - Duration of collisions Spells - RAVED

87        Here we go… We consider a molecule of mass m in a box of dimensions lx, ly and lz with a velocity components u1, v1 and w1 in the x, y and z directions respectively.

88 A derivation of two halves…
The strategy is to derive the pressure for one molecule on one face (1st half) and then sum all the pressures from all the molecules (2nd half). The first part involves Newton’s second law which states that the force on a body is equal to the rate of change of momentum.

89 Also note… The speed c of the molecule is given by:-
This comes from doing Pythagoras’ theorem in 3 dimensions. We will use this later in the 2nd half

90 Find the force on one molecule…
When the molecule impacts with Face A the x-component of its momentum is changed from mu1 to -mu1. But we want the rate of change of momentum, so need to divide by the time taken between collisions with Face A.

91 Time, t, between collisions…
So Newton’s 2nd Law gives us:- And Newton’s 3rd Law gives us:-

92 Pressure Recall that to work out the pressure:-
So the pressure p1 on face A:- So that’s the first bit done! – the pressure from one molecule in one direction.

93 Summing the pressures The total pressure on face A can be calculated by summing the pressures of all of the molecules:- Where p2, p3… refer to the pressures of all the other molecules up to N molecules.

94 Summing the pressures The last line can be rewritten as:-
Where the mean square x-component velocity is given by:- And similar equations can be derived for the y and z components

95 Almost there… But we want our equation to include the root mean square speed in all directions. Using Pythagoras’s theorem, the speed for one molecule is given by:- You can show that the root mean square speed for all the molecules is:-

96 Almost there… Because the motions are random we can write:-
Otherwise there would be a drift of particles in one direction. So using the above two equations:- So now we can write:-

97 Final Rearrangement The Kinetic Theory Equation:-
Can also be re-written as:- Because:-

98 Recap Use Newton’s Second Law Use Newton’s Third Law
Calculate pressure due to one molecule (force/area) Sum the pressures of all the molecules. Rewrite the speed in terms of root mean square speed.

99 Kinetic theory applets
There are a number of simulations on the internet. You will need to pick those that suit your specification and needs. Some suggestions are below: - The following were operational in October 2005 Boyle’s Law Effect of temperature and volume: Distribution of velocity: Molecular Model of an ideal gas:

100 Deriving Ideal Gas Equation
From Boyle’s Law: From Pressure Law: From Avogadro’s Law: Combining these three: Rewriting using the gas constant R: Therefore:-

101 Mean kinetic energy The mean kinetic energy of a molecule is the total kinetic energy of all the molecules/total number of molecules. The root mean square speed is defined as: …and so the mean kinetic energy of one molecule is gvien by…

102 Mean Kinetic Energy Note we have two equations with pV on the left hand side. (N=nNA) The Boltzmann constant, k, is defined as R/NA.

103 Mean Kinetic Energy Multiply both sides by 3/2.
Mean kinetic energy of a molecule of an ideal gas. Kinetic energy of one mole of gas (because R=NAk) Kinetic energy of n moles of gas.

104 Brownian Motion Thermal Physics

105 Homework Finish the past paper questions by next Friday (October 16th).

106 Learning Objectives Describe evidence supporting the kinetic theory of gases.

107 Robert Brown ( ) In 1827, the botanist Robert Brown was looking at pollen grains in water. He noticed that they constantly moved with a zigzag random motion.

108 Brownian Motion Experiment
We can observe Brownian motion in the lab. We put smoke in a brightly lit glass jar and observe the particles using a microscope. Observe bright specks moving haphazardly from side to side and up and down.

109 Brown couldn’t explain it…
Einstein used it 80 years later as evidence for the existence of molecules in the air. The randomly moving molecules were hitting the smoke particles unevenly, causing the motion.

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